Calculus Problem

Question

. Solve the equation 2sinx - radical 2 = 0. Give a general formula for all the solutions by using angle(s) in the interval *0,2π] and adding multiples of some integer k

Answer 1

2sinx - √2 = 0

 

sinx = √2/2      sinx is positive in quadrants 1 and 2

 

Solutions in [0, 2π]:  π/4 and 3π/4

 

sinx has period 2π, so adding or subtracting any multiple of 2π to either solution in the interval [0,2π] gives another solution.

 

Set of all solutions:

 

 x = π/4 + 2kπ or x = 3π/4 + 2kπ, where k =0, ±1, ±2,...   

Answer 2

2sinx - √2 = 0

 

 

Add √2 on both sides of the equation.

 

2sinx = √2

 

 

Divide both sides of the equation by 2.

 

sinx = √(2) / 2

 

 

x = π / 4

 

 

sine is positive in the 1st and 2nd quadrants.

 

x = kπ/4

 

where k = 1 and 3